题目

如图,点C为线段AB上一点,△ACM、△CBN是等边三角形.请你证明:(2)∠MFA=60?)△DEC为等边三角形

提问时间：2021-09-19

答案

∵∠NCB = 60?∵∠MAC = 60?∴MA∥NC ∴∠CEB=∠AMB ∵∠MCB = ∠MCN + ∠NCB = ∠MCN + 60?∵∠ACN = ∠MCN + ∠ACM = ∠MCN + 60?∴∠MCB = ∠ACN ∵△ACM、△CBN是等边三角形 ∴AC = MC、CN=CB ∴△ACN≌△MCB ∴∠NAC = ∠BMC ∵∠CDN=∠DCA + ∠NAC = 60?∠NAC = 60?∠BMC = ∠AMB = ∠CEB ∵∠ECB = 60?180?∠DCE - ∠ACD = 180?∠DCE - 60?120?∠DCE ∴ ∠DCE = 60?∠ECB ∵ CB = CN ∴△ECB≌△DCN ∴CD = CE ∴∠CED = ∠CDE ∵∠DCE = 60?∴∠CED = ∠CDE = (180?60?2 = 60?∴△DEC = 为等边三角形 ∴∠MFA = ∠FBA + ∠FAB = ∠ANC + ∠NAC = 180?(∠DCA + ∠DCN)= 180?120?∴∠MFA = 60?div>00